Electromagnetic Field as a Connection in a Vector Bundle

Question

I would like to know more about Ehresmann connections in vector bundles and how they relate to the electromagnetic field and the electron in quantum mechanics.

Background: The Schrödinger equation for a free electron is

$$ \frac{(-i\hbar\nabla)^2}{2m} \psi = i\hbar\partial_t \psi $$

Now, to write down the Schrödinger equation for an electron in an electromagnetic field given by the vector potential $A=(c\phi,\mathbf{A})$, we simply replace the momentum and time operator with the following operators

$$\begin{array}{rcl} -i\hbar\nabla &\mapsto& D_i = -i\hbar\nabla + e\mathbf{A} \\ i\hbar\partial_t &\mapsto& D_0 = i\hbar\partial_t - e\phi \end{array}$$

I have heard that this represents a "covariant derivative", and I would like to know more about that.

My questions:

1. (Delegated to Notation for Sections of Vector Bundles.)

2. I have heard that a connection is a "Lie-algebra-valued one-form". How can I visualize that? Why does it take values in the Lie-algebra of $U(1)$?

3. Since a connection is a one-form, how can I apply it to a section $\psi$? I mean, a one-form eats vectors, but I have a section here? What is $D_\mu \psi(x^\mu)$, is it a section, too?

I apologize for my apparent confusion, which is of course the reason for my questions.

Answer 1

Here's a short(-ish) answer. A vector bundle is a family of vector spaces over a manifold. The vector spaces can have bases. The manifold can have coordinates. The two concepts are not a priori related (now for the bundle of tangent spaces, a coordinate change happens to induce a change of basis; this fact often sows confusion). Once you pick a basis for your vector space, you do define a vector by its components -- but someone else may be describing the same vector in a different basis. To translate to physics: change of basis = gauge transformation.

In the case of a charged particle, the wave function is the component of a one-vector section; in a new basis, this number changes by a non-zero complex number (which can vary from point to point). Again, the wave function is a section, and a section means one vector for each point in the manifold. How do we differentiate a function which takes values in different vector spaces over different points? We need a way of connecting the vector spaces. The connection does this; pragmatically, it is just a rule for doing this differentiation.

Of course, differentiation will look different in different vector spaces, so the form of the connection will depend on the basis and change under gauge transformations (just as the form of a linear transformation changes under change of bases). That's what various messy formulas about how things "transform" are trying to tell you.

Answer 2

1. Not really sure about this question. Are you perhaps just asking about notation? You can choose whatever you like. But usually you choose one coordinate system and just work in that. It's certainly a non-issue if you are working in flat spacetime: there you have nice global coordinates for everything.

2. • First sentence is correct. For arbitrary Lie group $G$ with Lie algebra $\mathfrak{g}$ you might get a so called G-structure (e.g. $O(n)$-structure for Riemannian manifolds) and you can define a connection on that. It's rather heavy-duty mathematics, but at the end of the day you obtain a $\mathfrak{g} $-valued (more precisely, it takes values in the adjoint representation of $\mathfrak{g} $) one-form $ \omega $ and covariant derivative $D_{\mu}$ such as the one you've written.

   • In general I visualize connections like this: you put in a vector and the connection gives you back an element of the Lie algebra that is a generator of some transformation in the Lie group. E.g. on a Riemannian manifold you insert into the connection the direction you want to go to and you'll get back (very roughly said) information on how much the space is curved in that direction. More preciselly, if you integrate $\oint_{\gamma} \exp(\omega(\dot{\gamma}(t))) dt$ you obtain some element $T \in G$ that tells you how any vector gets parallel-transported along the closed curve $\gamma$ (this is called holonomy).

   • As for the last question of why $ U(1) $: well, because it's electromagnetism. Various groups will give you various interactions (e.g. $SU(3)$ gives you QCD). These groups arise because the theories contain something called gauge symmetry. I am sure you know that Maxwell equations, when written out in $A$ and $\phi$, are invariant to certain transformations. More concretely and clearly, as ${\bf F} = d{\bf A}$, where ${\bf F}$ is the electromagnetic tensor and ${\bf A} $ is the four-potential. In this form it's obvious that equations for $ {\bf F}$ don't change upon the transformation ${\bf A'} = {\bf A} + d\chi$. Now, let us step back to $\psi$ and note that if you want to make the theory locally invariant (why would you do that? because it's nice to have local properties instead of global ones. And the theory is clearly invariant with respect to global phase change, so let's just try this) with respect to the change of phase, you'll have to introduce new degree of freedom ${\bf A}$ that transforms precisely like the four-potential in Maxwell theory! Actually, what we just did is that we recovered photons. So, to conclude: if you choose $ U(1)$ as a group of symmetries, electromagnetism pops out. But note that to incorporate all of this consistently, you need to work in the framework of quantum field theory because Schrödinger's equation clearly isn't relativistically invariant, which is a feature we'd surely like to have in the theory of electromagnetism.

3. This can be a little confusing, especially in the physical literature and I doubt I can really make it clearer. Probably this will just confuse you even more but here goes: there exist various spaces of sections and $ G$-structures and representations of both $ G $ and $ \mathfrak{g}$ that one should really discern between but which are identified in physics. To make this rigorous would take too much space, so I suggest you look at some books on gauge theory. I'll just tell you that vector (more precisely vector field) is also a section and vice-versa. But note that another two terms are getting mixed together here: abstract vector as a concept from linear algebra (this is our $ \psi $: you know it's this kind of vector because it has no spacetime index $\mu $) and vector as an element of Minkowski spacetime, say $V^{\mu}$. As for the $ D_{\mu} $, it acts (by extension) also on tensor algebra of the space of sections $ \psi $ (in this simple case where $\psi $ has no index, the tensor algebra is isomorphic to normal tangent tensor algebra) and gives you a one-form living in this tensor algebra, so yeah: it is a section, but in a totally different (altough isomorphic) space!

No need to apologize. These things are pretty hard and it takes a lot of time to grok all of it and I guess lot of the physicists would just hand-wave most of the mathematical content away and go ahead to calculate something. Which is possible here because $ U(1) $ is only one-dimensional, its Lie algebra is one-dimensional, and so is $ \psi $. So you just work with numbers all the time and there arises no need for abstract concepts. Except that this doesn't help you one bit when you'll try to generalize this (either to QCD with non-abelian group $ SU(3) $, or to the curved spacetime), or even if you try to think about some concepts that are rather basic from the point of view of geometry (e.g. try thinking about what is the meaning of the above-mentioned holonomy with respect to $\gamma $ in the case of $ U(1 ) $). So it's good you're trying to understand the core principles already.

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Update in regard to Greg's question in the comments:

I am not sure I understand you completely but I got a feeling (probably wrong) that you are mixing up various notions of invariance here. There are at least two notions of physical invariance (under the action of Lorentz group and under the action of $U(1)$) and also a notion of invariance in the sense of being "coordinate-free". Now, if my feeling is correct, you are asking for an analogue of $v = v^{\mu}{{\rm \partial} \over {\rm \partial} x^{\mu}}$ for $\psi$. These two things are indeed very similar, but in a little disguised way. More precisely: $v$ is a section of a tangent bundle $TM$ and we are decomposing it with respect to some section of the canonical tangent frame bundle $FM$, which also carries a natural action of the group $GL_k({\mathbb{R}})$ (the action is a local change of basis and $k$ is the rank of $TM$). In other words, we have a $G$-structure here and this is where "coordinate-free" invariance comes from. The situation is similar with $\psi$: it is a section of a vector bundle $\pi: V \to M$ which carries an $U(1)$-structure. At this point it should also be clear where the difference between the two cases is: in the former you have two bundles $TM$ and $FM$ while in the latter there is only $\pi: V \to M$. So it doesn't really make sense to ask for $\psi$ to be any more invariant than it already is: you have nothing with respect to which you could decompose it. So instead of thinking about $\psi$ as an analogue of section of $TM$, think of it instead as an analogue of a section of $FM$.

Answer 3

I can fully understand your confusion since it is natural that you feel overwhelmed by this new viewpoint on the theory.

The answers given by Eric and Marek are just fine and I will not directly talk about principal bundles, local trivialization and the like. I want to present a very intuitive approach here.

I would suggest that you go one or two steps back and try to understand the notion of a covariant derivative in classical differential geometry. There, the covariant derivative $D$ assures that if you derive some quantity $F$ on a manifold, say some surface, this new quantity $DF$ will also lie "on the manifold" (actually, something related to it like the tangent space).

The following example will hopefully illustrate the issue what it means that something has to "stay on the manifold".

Mass-point on a surface

Ok, lets do the most simple example one could think of, the motion of a free mass point on a surface in Newtonian mechanics. As you know, the Lagrangian in this case is just the kinetic energy,

$$L = T = \frac{m}{2}\mathbf{v}^2$$

So, what is now $\mathbf{v}^2$? We have to assume that in every point, the velocity is tangential to the surface. Then, we know that $\mathbf{v}^2 = g_{ab}\dot{x}^a\dot{x}^b$ where we sum over the indices and the surface is described by some metric $g_{ab}(x)dx^adx^b$ and we have replaced the velocity by the time derivative of the position of the particle.

For the solution of the system we will need two terms. First of all, we want to calculate

$$\frac{\partial{L}}{\partial{x^k}} = \frac{m}{2}\partial_k{g_{ab}(x)}\dot{x}^a\dot{x}^b$$

second,

$$\frac{d}{dt}\frac{\partial{L}}{\partial{\dot{x}^k}} = \frac{d}{dt}\frac{m}{2}g_{ab}(x) \left( \delta^a_k\dot{x}^b + \dot{x}^a\delta^b_k \right) = m\frac{d}{dt}g_{kb}(x)\dot{x}^b$$

since $g$ is symmetric. Now,

$$m\frac{d}{dt}g_{kb}(x)\dot{x}^b = m\left( \partial_lg_{kb}(x)\dot{x}^l\dot{x}^b + g_{kb}(x)\ddot{x}^b\right) $$

Finally, the equations of motion are given by

$$\frac{d}{dt}\frac{\partial{L}}{\partial{\dot{x}^k}} - \frac{\partial{L}}{\partial{x^k}} = 0$$

and dropping m, re-using the symmetry in $g$ and renaming some indices, we arrive at

$$g_{kb}\ddot{x}^b+\frac{1}{2}\left( \partial_a g_{kb} + \partial_b g_{ka} - \partial_k g_{ab} \right)\dot{x}^a\dot{x}^b = 0$$

which is exactly by applying $g^{ik}$

$$\ddot{x}^i + \Gamma^i_{ab}\dot{x}^a\dot{x}^b = 0$$

Where the Christoffel symbols can directly seen as

$$\Gamma^i_{ab} = \frac{1}{2}g^{ik}\left( \partial_a g_{kb} + \partial_b g_{ka} - \partial_k g_{ab} \right)$$

and I hope I did not miscalculate anything.

Relation to electrodynamics

Now this equation of motion is already magic since it is precisely the equation of motion for a testparticle in general relativity. But what about (other) gauge field theories?
Here, the curvature is not defined with respect to the manifold directly but to a group somehow "attached" to it. That's why it will have some group indices but one can drop this if the Lie Algebra of the group is one dimensional as in the case of electrodynamics. There, our curvature is $F_{\mu\nu}$ but we could also state $F^a_{\mu b\nu}$ where now $a$ and $b$ are indices of the group. This looks much more familliar to the curvature of general relativity, $R^{\mu}_{\nu\alpha\beta}$ where all indices correspond to the manifold, in some sense, the tangent space is the group of general relativity, roughly speaking.

For historical reasons, the Christoffel symbols that somehow catch (not invariantly!) the, force, acting on the particle because of the curvature are in gauge theories called gauge fields $A$ and rescaled,

$$\beta A^a_{\mu b}dx^{\mu}" = "\Gamma^a_{\mu b}dx^\mu $$

with again group indices $a$ and $b$.

Now if you derive something on the manifold, you will always have to define that derivative with respect to the Christoffel Symbols to stay "in the manifold". On the other hand, the derivation will also have to respect the group character, one could say the result has to stay "in the group". This will be realized by a covariant derivative and here the "Christoffel symbols" are called gauge fields.

Sincerely

Robert

Answer 4

The best book I've seen on this is by Chris Isham called Modern Differential Geometry for Physicists. It's heavy going because he does everything rigorously but it is worth it in the end for conceptual clarification.

There's basically two concepts that one has to get a hold of: principal bundles and vector bundles. The notion of a connection can be defined in both independently and it's useful to know how to translate between the two.

A principal bundle has a structure group, generally known in physics as the gauge group, and the main construction that gets you a vector bundle from this is called the associated bundle which is obtained by supplying a representation of the structure/gauge group.

1) The notation is heavy because there is a mass of detail to keep a hold of. Each step is straight-forward but it's the sheer amount of it.

2) A connection on a principal bundle $E$ is a splitting of its tangent bundle $TE$ into a vertical bundle $VE$ and a horizontal bundle $HE$ that is equivariant, this means its compatible with the action of the structure group on the fibres. The connection 1-form arises from observing that we can identify the fibres of the vertical bundle with the tangent bundle of the group $G$ at the origin $e$, that is we get $T_eG$, and this is exactly the Lie algebra of the group.

3) One doesn't use the connection 1-form directly in the way you've indicated. Instead, a covariant derivative is constructed from connection 1-form and this itself is not a 1-form.

It's worth pointing out that principal and vector bundles and the connections on them are global constructions, whereas the constructions that one usually sees in physics are typically local. To get the local picture one usually takes a section of the bundle and then pullback the geometrical structure.

This was a point that confused me. For example, the mathematical connection 1-form is global, but the physicists connection 1-form is local and they are different.